Question: $x^3y+y^2-x^2=5$ Find the value of $\dfrac{dy}{dx}$ at the point $(2,1)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $\dfrac{1}{5}$ (Choice C) C $-\dfrac{4}{5}$ (Choice D) D $\dfrac27$
Explanation: We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $x^3y+y^2-x^2=5$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} x^3y+y^2-x^2&=5 \\\\ \dfrac{d}{dx}(x^3y+y^2-x^2)&=\dfrac{d}{dx}(5) \\\\ \dfrac{d}{dx}(x^3y)+\dfrac{d}{dx}(y^2)-\dfrac{d}{dx}(x^2)&=0 \\\\ 3x^2\cdot y+x^3\cdot\dfrac{dy}{dx}+2y\cdot\dfrac{dy}{dx}-2x&=0 \\\\ 3x^2y+x^3\cdot\dfrac{dy}{dx}+2y\cdot\dfrac{dy}{dx}-2x&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 3x^2y+x^3\cdot\dfrac{dy}{dx}+2y\cdot\dfrac{dy}{dx}-2x&=0 \\\\ \dfrac{dy}{dx}(x^3+2y)&=2x-3x^2y \\\\ \dfrac{dy}{dx}&=\dfrac{2x-3x^2y}{x^3+2y} \end{aligned}$ Now we can plug the point $(2,1)$ into the expression for $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{2x-3x^2y}{x^3+2y} \\\\ &=\dfrac{2(2)-3(2)^2(1)}{(2)^3+2(1)} \gray{x=2,\,\,y=1} \\\\ &=\dfrac{-8}{10} \\\\ &=-\dfrac{4}{5} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at the point $(2,1)$ is $-\dfrac{4}{5}$.